# How to replace the elements of a matrix using the conditions if,else?

370 visualizaciones (últimos 30 días)
Emerson De Souza el 5 de Jun. de 2011
I want to replace the elements of a matrix
using different conditions. For instance, let all
elements larger than 0.5 be replaced by -1, else
keep the way it is.
So I thought I should simply write the command below:
X=rand(10,10);
if X(:,:)>0.5;
T(:,:)=-1;
else T=X(:,:);
end;
but it does not work because T==X.
Could someone tell me how to correct these command lines?
Thank you
Emerson
##### 0 comentariosMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos

Iniciar sesión para comentar.

Matt Fig el 5 de Jun. de 2011
IF statements do not pick out elements of an array like you are imagining that they do. When you write this:
if conditional
% Do something
end
for non-scalar conditional, the IF statement will pass if and only if all of the elements in conditional are true or non-zero. For example:
x = [1 2 3]; % All are non-zero, passes conditional...
if x
disp('In if')
end
but now change it to:
x = [1 2 0]; % All are not non-zero, fails conditional...
if x
disp('In if')
end
So the way you have to do what you want is either through logical indexing, or by single value through iteration.
% iteration approach - look at one value at a time!
x = [.1 .2 .3 .7 .8 .9 .1 .2 .3]; % Work with this array.
T = zeros(size(x)); % Make another array to fill up...
for ii = 1:length(x)
if x(ii)>.5
T(ii) = 99;
else
T(ii) = -100;
end
end
% Logical indexing approach
x = [.1 .2 .3 .7 .8 .9 .1 .2 .3]; % Work with this array.
T(x>.5) = 99;
T(x<=.5) = -100
##### 4 comentariosMostrar 3 comentarios más antiguosOcultar 3 comentarios más antiguos
Nick el 30 de Oct. de 2019
Mattfig,
What should be done if each individual element must be sorted through but there are thousands of elements. My script takes way too long to run. Is there any optimiziation?

Iniciar sesión para comentar.

### Más respuestas (6)

Rain el 11 de Dic. de 2013
Hi,
Actually, there is a very simple way to do it:
X=rand(10,10);
X(X>0.5) = [-1];
##### 4 comentariosMostrar 3 comentarios más antiguosOcultar 3 comentarios más antiguos
Majid Mahmoodabadi el 16 de Abr. de 2022
Hi,
I have a matrix as following and want to increase the value of numbers (by 1) that are less than 9.
B= [12,16,5,6,7,11,13,22]
I applied the below logic:
B(B<9)= [B(i)+1]
This is what I got:
B =
12 16 17 17 17 11 13 22
But, I look for a response like this:
B =
12 16 6 7 8 11 13 22
I appreciate it if you could help me.

Iniciar sesión para comentar.

Ivan van der Kroon el 5 de Jun. de 2011
You don;t need the if-statment here but only the logicals. This gives you a matrix with ones where X is larger than 0.5 and zeros other wise
(X>0.5)
T = X.* (X<=0.5)-(X>0.5);
Alls values smaller than or equal to 0.5 are kept while the others are set to zero and then a matrix is added that has entries of -1 for the entries of X larger than 0.5.
##### 1 comentarioMostrar NingunoOcultar Ninguno
Emerson De Souza el 5 de Jun. de 2011
Hi Ivan
For this particular case you are right.
But how do we write, if the matrix T is build up
based on conditions of several matrices?
For example: The elements of T equals -1 if:
1) the elements of L>0.5 AND
2) the elements of M=0 AND
3) the elements of N<0.5
if any of these conditions is not satisfied, T equals the elements of M+1
Hope there is a way to do it
Emerson

Iniciar sesión para comentar.

Ivan van der Kroon el 5 de Jun. de 2011
Just implement it for multiple matrices using element multplication:
Logical=(L>0.5).*(M==0).*(N<0.5);
T=-Logical+(M+1).*Logical;
##### 3 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Kelly Kearney el 11 de Dic. de 2013
That seems a little convoluted (I see how you're combining both assignments into one, but for a beginner the syntax might not be clear). This might be better:
T = M + 1;
T(L > 0.5 & M == 0 & N < 0.5) = -1;

Iniciar sesión para comentar.

yashar khatib shahidi el 3 de Mayo de 2015
I have a row vector like a = [3 4 6 8 9] then I want to replace third element which is 6 with 5 and 8 so the new vector becomes b = [3 4 5 8 8 9]. What is the function to convert the vector. Thanks
##### 0 comentariosMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos

Iniciar sesión para comentar.

Jyahway Dong el 19 de Oct. de 2016
This is very important message for me, spend hours try to debug this and thank you all
##### 0 comentariosMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos

Iniciar sesión para comentar.

Kinga Gyeltshen el 16 de Mzo. de 2021
I have a 9x9 matrix and in every iteration i want to retain the 3x3 matrix and on to it I want to add the 4,5,6,7,8,9 (row, column) element to the 3x3 matrix to form 4x4 matrix. The fourth element (row,column) should get replaced with the remaining element of the 9x9 matrix. Can anyone help me with a simple and condensed algorithm to get it done please.
Thank you.
##### 0 comentariosMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos

Iniciar sesión para comentar.

### Categorías

Más información sobre Introduction to Installation and Licensing en Help Center y File Exchange.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!